python equivalent of R table

Solution 1:

Pandas has a built-in function called value_counts().

Example: if your DataFrame has a column with values as 0's and 1's, and you want to count the total frequencies for each of them, then simply use this:

df.colName.value_counts()

Solution 2:

A Counter object from the collections library will function like that.

from collections import Counter

x = [[12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [6, 0], [12, 6], [0, 6], [12, 0], [0, 6], [0, 6], [12, 0], [0, 6], [6, 0], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [0, 6], [0, 6], [12, 6], [6, 0], [6, 0], [12, 6], [12, 0], [12, 0], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 0], [12, 0], [12, 0], [12, 0], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [12, 6], [12, 0], [0, 6], [6, 0], [12, 0], [0, 6], [12, 6], [12, 6], [0, 6], [12, 0], [6, 0], [6, 0], [12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [12, 0], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [0, 6], [12, 0], [12, 6], [0, 6], [0, 6], [12, 0], [0, 6], [12, 6], [6, 0], [12, 6], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [12, 6], [12, 0], [6, 0], [12, 6], [6, 0], [12, 0], [6, 0], [12, 0], [6, 0], [6, 0]]

# Since the elements passed to a `Counter` must be hashable, we have to change the lists to tuples.
x = [tuple(element) for element in x]

freq = Counter(x)

print freq[(12,6)]

# Result:  28

Solution 3:

Supposing you need to convert the data to a pandas DataFrame anyway, so that you have

L = [[12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [6, 0], [12, 6], [0, 6], [12, 0], [0, 6], [0, 6], [12, 0], [0, 6], [6, 0], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [0, 6], [0, 6], [12, 6], [6, 0], [6, 0], [12, 6], [12, 0], [12, 0], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 0], [12, 0], [12, 0], [12, 0], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [12, 6], [12, 0], [0, 6], [6, 0], [12, 0], [0, 6], [12, 6], [12, 6], [0, 6], [12, 0], [6, 0], [6, 0], [12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [12, 0], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [0, 6], [12, 0], [12, 6], [0, 6], [0, 6], [12, 0], [0, 6], [12, 6], [6, 0], [12, 6], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [12, 6], [12, 0], [6, 0], [12, 6], [6, 0], [12, 0], [6, 0], [12, 0], [6, 0], [6, 0]]
df = pd.DataFrame(L, columns=('a', 'b'))

then you can do as suggested in this answer, using groupby.size():

tab = df.groupby(['a', 'b']).size()

tab looks as follows:

In [5]: tab
Out[5]:
a   b
0   6    19
6   0    20
12  0    33
    6    28
dtype: int64

and can easily be changed to a table form with unstack():

In [6]: tab.unstack()
Out[6]:
b      0     6
a
0    NaN  19.0
6   20.0   NaN
12  33.0  28.0

Fill NaNs and convert to int at your own leisure!

Solution 4:

import pandas
x = [[12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [6, 0], [12, 6], [0, 6], [12, 0], [0, 6], [0, 6], [12, 0], [0, 6], [6, 0], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [0, 6], [0, 6], [12, 6], [6, 0], [6, 0], [12, 6], [12, 0], [12, 0], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 0], [12, 0], [12, 0], [12, 0], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [0, 6], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [12, 6], [12, 0], [0, 6], [6, 0], [12, 0], [0, 6], [12, 6], [12, 6], [0, 6], [12, 0], [6, 0], [6, 0], [12, 6], [12, 0], [0, 6], [12, 0], [12, 0], [12, 0], [6, 0], [12, 6], [12, 6], [12, 6], [12, 6], [0, 6], [12, 0], [12, 6], [0, 6], [0, 6], [12, 0], [0, 6], [12, 6], [6, 0], [12, 6], [12, 6], [12, 0], [12, 0], [12, 6], [0, 6], [6, 0], [12, 0], [6, 0], [12, 0], [12, 0], [12, 6], [12, 0], [6, 0], [12, 6], [6, 0], [12, 0], [6, 0], [12, 0], [6, 0], [6, 0]] 
ps = pandas.Series([tuple(i) for i in x])
counts = ps.value_counts()
print counts

you will get the result like:

(12, 0)    33
(12, 6)    28
(6, 0)     20
(0, 6)     19

and for [(12,6)] you will get exact number, here 28

more about pandas, which is powerful Python data analysis toolkit, you can read in official doc: http://pandas.pydata.org/pandas-docs/stable/

UPDATE:

If order does not matter just use sorted: ps = pandas.Series([tuple(sorted(i)) for i in x]) after that result is:

(0, 6)     39
(0, 12)    33
(6, 12)    28

Solution 5:

IMHO, pandas offers a better solution for this "tabulation" problem:

One dimension:

my_tab = pd.crosstab(index = df["feature_you_r_interested_in"],
                              columns="count")

Proportion count:

my_tab/my_tab.sum()

Two-dimensions (with totals):

cross = pd.crosstab(index=df["feat1"], 
                             columns=df["feat2"],
                             margins=True)

cross

Also, as mentioned by other coleagues, pandas value_counts method could be all you need. It is so good that you can have the counts as percentages if you want:

df['your feature'].value_counts(normalize=True)

I'm very grateful for this blog:

http://hamelg.blogspot.com.br/2015/11/python-for-data-analysis-part-19_17.html